Given an array arr[] of n positive integers. The task is to find the size of the subset formed from the elements of the given array and the absolute difference between any two elements of the set is less than equal to 1.
Examples :
Input : arr[] = {8, 9, 8, 7, 8, 9, 10, 11}
Output : 5
If we make subset with elements {8, 9, 8, 8, 9}.
Each pair in the subset has an absolute
difference <= 1
Input : arr[] = {4, 5, 2, 4, 4, 4}
Output : 5
Subset is {4, 5, 4, 4, 4}
Observe, since we want the absolute difference between any two elements to be less than equal to 1, then there can be a maximum of two distinct numbers. So, the subset we choose will be in the form {a, a, a, ….., b, b, b} or {a, a, a, a, …..}.
Now, to find the size of such subset we will find the frequency of each element say c1, c2, c3, …., cj, …., cmaximum element in arr. Then our answer will be the maximal value of ci + ci+1.
Below is the implementation of this approach:
C++
// CPP Program to find the size of// the subset formed from the elements // of the given array such that the // maximum difference is 1#include <bits/stdc++.h>using namespace std;// Return the maximum size of subset with // absolute difference between any element // is less than 1.int maxsizeSubset(int arr[], int n){ // Inserting elements and their // frequencies in a hash table. unordered_map<int, int> mp; for (int i = 0; i < n; i++) mp[arr[i]]++; // Traverse through map, for every element // x in map, find if x+1 also exists in map. // If exists, see if sum of their frequencies // is more than current result. int res = 0; for (auto x : mp) if (mp.find(x.first + 1) != mp.end()) { res = max(res, mp[x.first] + mp[x.first+1]); } else { res=max(res,mp[x.first]); } return res;}// Driven Programint main(){ int arr[] = {1, 2, 2, 3, 1, 2}; int n = sizeof(arr) / sizeof(arr[0]); cout << maxsizeSubset(arr, n) << endl; return 0;} |
Java
// Java Program to find the size of // the subset formed from the elements // of the given array such that the // maximum difference is 1 import java.util.*;class GFG { // Return the maximum size of subset with // absolute difference between any element // is less than 1. static int maxsizeSubset(int arr[], int n) { // Inserting elements and their // frequencies in a hash table. Map<Integer, Integer> mp = new HashMap<>(); for (int i = 0; i < n; i++) { if (mp.containsKey(arr[i])) { mp.put(arr[i], mp.get(arr[i]) + 1); } else { mp.put(arr[i], 1); } } // Traverse through map, for every element // x in map, find if x+1 also exists in map. // If exists, see if sum of their frequencies // is more than current result. int res = 0; for (Map.Entry<Integer, Integer> x : mp.entrySet()) { if (mp.containsKey(x.getKey() + 1)) { res = Math.max(res, mp.get(x.getKey()) + mp.get(x.getKey() + 1)); } else { res = Math.max(res, mp.get(x.getKey())); } } return res; } // Driver code public static void main(String[] args) { int arr[] = {1, 2, 2, 3, 1, 2}; int n = arr.length; System.out.println(maxsizeSubset(arr, n)); }}/* This code is contributed by PrinciRaj1992 */ |
Python3
# Python Program to find the size of# the subset formed from the elements# of the given array such that the# maximum difference is 1# Return the maximum size of subset with# absolute difference between any element# is less than 1.def maxsizeSubset(arr, n): # Inserting elements and their # frequencies in a hash table. mp = {} for i in range(n): if (arr[i] in mp): mp[arr[i]] = mp[arr[i]] + 1 else: mp[arr[i]] = 1 # Traverse through map, for every element # x in map, find if x+1 also exists in map. # If exists, see if sum of their frequencies # is more than current result. res = 0 for x,y in mp.items(): if ((x + 1) in mp): res = max(res, mp[x] + mp[x + 1]) else: res = max(res, mp[x]) return res# Driven Programarr = [1, 2, 2, 3, 1, 2]n = len(arr)print(maxsizeSubset(arr, n))# This code is contributed by Shinjanpatra |
C#
// C# Program to find the size of // the subset formed from the elements // of the given array such that the // maximum difference is 1 using System;using System.Collections.Generic;class GFG { // Return the maximum size of subset with // absolute difference between any element // is less than 1. static int maxsizeSubset(int []arr, int n) { // Inserting elements and their // frequencies in a hash table. Dictionary<int, int> mp = new Dictionary<int, int>(); for (int i = 0 ; i < n; i++) { if(mp.ContainsKey(arr[i])) { var val = mp[arr[i]]; mp.Remove(arr[i]); mp.Add(arr[i], val + 1); } else { mp.Add(arr[i], 1); } } // Traverse through map, for every element // x in map, find if x+1 also exists in map. // If exists, see if sum of their frequencies // is more than current result. int res = 0; foreach(KeyValuePair<int, int > x in mp) { if (mp.ContainsKey(x.Key + 1)) { res = Math.Max(res, mp[x.Key] + mp[x.Key + 1]); } else { res = Math.Max(res, mp[x.Key]); } } return res; } // Driver code public static void Main(String[] args) { int []arr = {1, 2, 2, 3, 1, 2}; int n = arr.Length; Console.WriteLine(maxsizeSubset(arr, n)); } } // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript Program to find the size of// the subset formed from the elements// of the given array such that the// maximum difference is 1// Return the maximum size of subset with// absolute difference between any element// is less than 1.function maxsizeSubset(arr, n) { // Inserting elements and their // frequencies in a hash table. let mp = new Map(); for (let i = 0; i < n; i++) { if (mp.has(arr[i])) { mp.set(arr[i], mp.get(arr[i]) + 1) } else { mp.set(arr[i], 1) } } // Traverse through map, for every element // x in map, find if x+1 also exists in map. // If exists, see if sum of their frequencies // is more than current result. let res = 0; for (let x of mp) if (mp.has(x[0] + 1)) { res = Math.max(res, mp.get(x[0]) + mp.get(x[0] + 1)); } else { res = Math.max(res, mp.get(x[0])); } return res;}// Driven Programlet arr = [1, 2, 2, 3, 1, 2];let n = arr.length;document.write(maxsizeSubset(arr, n) + "<br>");// This code is contributed by gfgking.</script> |
Output
5
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space : O(n)
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